Problem: Graph this system of equations and solve. $y = -\dfrac{7}{2} x + 4$ $y = -\dfrac{1}{2} x - 2$ 1 2 3 4 5 6 7 8 9 10 \llap{-}2 \llap{-}3 \llap{-}4 \llap{-}5 \llap{-}6 \llap{-}7 \llap{-}8 \llap{-}9 \llap{-}10 1 2 3 4 5 6 7 8 9 10 \llap{-}2 \llap{-}3 \llap{-}4 \llap{-}5 \llap{-}6 \llap{-}7 \llap{-}8 \llap{-}9 \llap{-}10 Click and drag the points to move the lines.
The y-intercept for the first equation is $4$ , so the first line must pass through the point $(0, 4)$ The slope for the first equation is $-\dfrac{7}{2}$ . Remember that the slope tells you rise over run. So in this case for every $7$ positions you move down (because it's negative) You must also move $2$ positions to the right. $2$ positions to the right. $7$ positions down from $(0, 4)$ is $(2, -3)$ Graph the blue line so it passes through $(0, 4)$ and $(2, -3)$ The y-intercept for the second equation is $-2$ , so the second line must pass through the point $(0, -2)$ The slope for the second equation is $-\dfrac{1}{2}$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move down (because it's negative) You must also move $2$ positions to the right. $2$ positions to the right. Graph the green line so it passes through $(0, -2)$ and $(2, -3)$ The solution is the point where the two lines intersect. The lines intersect at $(2, -3)$.